Semiconductors

JEE Main level, single correct, medium

134 of 209

Question 134 of 209

For the given logic gate circuit, which of the following is the correct truth table?

Show answer and solution

Answer: (D) \(\begin{array}{c|c|c} n & m & z \\ \hline 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{array}\)

The given logic circuit consists of an OR gate and a NAND gate.
The output of the OR gate is y=n+my = n + m.
The inputs to the NAND gate are nn and yy.
The final output zz is given by z=n⋅y‾=n⋅(n+m)‾z = \overline{n \cdot y} = \overline{n \cdot (n + m)}.
Using Boolean algebra: n⋅(n+m)=n⋅n+n⋅m=n+n⋅m=n(1+m)=nn \cdot (n + m) = n \cdot n + n \cdot m = n + n \cdot m = n(1 + m) = n.
Therefore, z=n‾z = \overline{n}.
Let's verify this with the truth table:
nmy = n + mz = NAND(n, y)0001011111101010
Comparing this with the given options, option (4) matches the calculated truth table.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Semiconductors, diodes and logic gates simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Semiconductors (Logic gates), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.